Gardens and Fencing

Looking at the graph, we know that $\triangle A'B'D'$ is an equilateral triangle $\implies$ $\angle D'A'B'= \angle A'B'D'= \angle A'D'B'= 60^{\circ}$ .

We know that each of the line segments is tangent to the graph; thus $quadrilaterals \ ABB'A', A'D'EF, and D'B'CD$ are all rectangles,

$|AB| = |CD|= | EF|= 2\times r= 2\times 9= 18 \implies$ the measure of the three segments is $\ 3\times 18= 54 units$ ;

$\implies$ $\angle AA'B'= \angle BB'A'= \angle FA'D'= \angle A'D'E=\angle DD'B'=\angle D'B'C= 90^{\circ}$ .

$\implies$ $\angle AA'F= \angle DD'E= \angle CC'F= 120^{\circ}$

$\implies$ $arc{AF}= arc{BC}= arc{DE}= 2\pi r\times \frac{120}{360}= 6\pi$ ,

Thus, the length of the fence is $18\pi+ 54 \implies x= 54 \ and\ y= 18 \implies \frac{18}{3}+54= \boxed{60}$

Note: The graph is not perfectly drawn to scale, my internet was too slow but you got the idea.

Just visualize that the three arcs together forms one circle of the same radius that is 9cm and the line segments between are just the projections of the equilateral triangle of side 18cm formed inside . The length of the fence is therefore 2πr + 3x18 i.e, 18π+54 . Hence , y/3+x = 18/3+54 = 6+54= 60. Absolutely no need of even picking up the pen and the notebook. 😉